Đáp án:
$\begin{array}{l}
5)c)4\left( {{x^2} - \dfrac{1}{2}y} \right)\left( {{x^2} + \dfrac{1}{2}y} \right)\\
= 4.\left( {{x^4} - \dfrac{1}{4}{y^2}} \right)\\
= 4{x^4} - {y^2}\\
d)\left( {x - 2} \right)\left( {3 - {x^2}} \right)\left( {x + 1} \right)\\
= \left( {3x - {x^3} - 6 + 2{x^2}} \right)\left( {x + 1} \right)\\
= \left( { - {x^3} + 2{x^2} + 3x - 6} \right)\left( {x + 1} \right)\\
= - {x^4} + 2{x^3} + 3{x^2} - 6x\\
- {x^3} + 2{x^2} + 3x - 6\\
= - {x^4} + {x^3} + 5{x^2} - 3x - 6\\
B8)\\
\left( {xy + 1} \right)\left( {{x^2}{y^2} - xy + 1} \right) + \left( {{x^3} - 1} \right)\left( {1 - {y^3}} \right)\\
= {\left( {xy} \right)^3} + {1^3} + {x^3} - {x^3}{y^3} - 1 + {y^3}\\
= {x^3}{y^3} + 1 + {x^3} - {x^3}{y^3} - 1 + {y^3}\\
= {x^3} + {y^3}
\end{array}$
